Post on 04-Jan-2016
UBC March 2007 1
The Evergreen Project: The Promise of Polynomials to Boost CSP/SAT Solvers*
Karl J. Lieberherr
Northeastern University
Boston
joint work with Ahmed Abdelmeged, Christine Hang and Daniel Rinehart
Title inspired by a paper by Carla Gomes / David Shmoys
UBC March 2007 2
Two objectives
• I want you to become – better writers of MAX-SAT/MAX-CSP solvers
• better decision making• crosscutting exploration of search space
– better players of the Evergreen game• the game reveals what the polynomials can do• iterated game of base zero-sum game:
– Together: choose domain– Anna: choose instance (min .max. possible loss)– Bob: solve instance, Anna pays Bob satisfaction fraction
• perfect information game
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Introduction
• Boolean MAX-CSP(G) for rank d, G = set of relations of rank d– Input
• Input = Bag of Constraint = CSP(G) instance• Constraint = Relation + Set of Variable• Relation = int. // Relation number < 2 ^ (2 ^ d) in G• Variable = int
– Output• (0,1) assignment to variables which maximizes the number of
satisfied constraints.
• Example Input: G = {22} of rank 3. H =– 22:1 2 3 0 – 22:1 2 4 0 – 22:1 3 4 0 1in3 has number 22
M = {1 !2 !3 !4} satisfies all
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Variation
MAX-CSP(G,f): Given a CSP(G) instance H expressed in n variables which
may assume only the values 0 or 1, find an assignment to the n variables which satisfies at least the fraction f of the constraints in H.
Example: G = {22} of rank 3MAX-CSP({22},f): H =
22:1 2 3 0 22:1 2 4 0 in MAX-CSP({22},?). Highest value for ?22:1 3 4 022: 2 3 4 0 1in3 has number 22
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Evergreen(3,2) game
Anna and Bob: They agree on a protocol P1 to choose a set of 2 relations (=G) of rank 3.– Anna chooses CSP(G)-instance H (limited).– Bob solves H and gets paid by Anna the fraction
that Bob satisfies in H.• This gives nice control to Anna. Anna will choose an
instance that will minimize Bob’s profit.
– Take turns.R1, Anna
R2, Bob
100% R1, 0% R2100% R2, 0% R1
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Protocol choice
• Randomly choose R1 and R2 (independently) between 1 and 255 (Throw two dice choosing relations).
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Tell me
• How would you react as Anna?– The relations 22 and 22 have been chosen.– You must create a CSP({22}) instance with 1000
variables in which only the smallest possible fraction can be satisfied.
– What kind of instance will this be?
• What kind of algorithm should Bob use to maximize its payoff?
• Should any MAX-CSP solver be able to maximize Bob’s profit? How well does MAX-SAT (e.g., yices,
ubcsat)or MAX-CSP solvers do on symmetric instances ???
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Game strategy in a nutshell
• Choose G={R1,R2} randomly.• Anna chooses instance so that payoff is
minimized.• Bob finds solution so that payoff is
maximized (Solve MAX-CSP(G))
• Take turns: Choose G= … Bob chooses …• Requires thorough understanding of MAX-
CSP(G) problem domain. Requires an excellent MAX-CSP(G) solver.
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Our approach by Example:SAT Rank 2 example
14 : 1 2 014 : 3 4 014 : 5 6 0 7 : 1 3 0 7 : 1 5 0 7 : 3 5 0 7 : 2 4 0 7 : 2 6 0 7 : 4 6 0
14: 1 2 = or(1 2) 7: 1 3 = or(!1 !3)
= H
Evergreen game:maximize payofffind maximum assignment
UBC March 2007 10appmean = approximation of the mean (k variables true)
Blurry vision
What do we learn from the abstract representation absH?• set 1/3 of the variables to true (maximize).• the best assignment will satisfy at least 7/9 constraints.• very useful but the vision is blurry in the “middle”.
excellent peripheral vision
0 1 2 3 4 5 6 = k
8/9
7/9
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Our approach by Example
• Given a CSP(G)-instance H and an assignment N which satisfies fraction f in H– Is there an assignment that satisfies more than f?
• YES (we are done), absH(mb) > f
• MAYBE, The closer absH() comes to f, the better
– Is it worthwhile to set a certain literal k to 1 so that we can reach an assignment which satisfies more than f
• YES (we are done), H1 = Hk=1, absH1(mb1) > f
• MAYBE, the closer absH1(mb1) comes to f, the better
• NO, UP or clause learningabsH= abstract representation of H
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14 : 1 2 014 : 3 4 014 : 5 6 0 7 : 1 3 0 7 : 1 5 0 7 : 3 5 0 7 : 2 4 0 7 : 2 6 0 7 : 4 6 0
14 : 2 014 : 3 4 014 : 5 6 0 7 : 1 3 0 7 : 1 5 0 7 : 3 5 0 7 : 2 4 0 7 : 2 6 0 7 : 4 6 0
8/9
6/7 = 8/9
3/7=5/9
3/9H
H0
abstract representation
0 1 2 3 4 5 6
0 1 2 3 4 5
maximum assignment away from max bias: blurry
7/9
5/7=7/9
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14 : 1 2 014 : 3 4 014 : 5 6 0 7 : 1 3 0 7 : 1 5 0 7 : 3 5 0 7 : 2 4 0 7 : 2 6 0 7 : 4 6 0
14 : 1 2 014 : 3 4 014 : 5 6 0 7 : 3 0 7 : 5 0 7 : 3 5 0 7 : 2 4 0 7 : 2 6 0 7 : 4 6 0
8/9
7/8=8/9
6/8=7/9
H
H1
3/8
2/7=3/8
0 1 2 3 4 5 6
0 1 2 3 4 5
maximum assignment away from max bias: blurry
7/9
clearlyabove 3/4
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8/97/9
14 : 1 2 014 : 3 4 014 : 5 6 0 7 : 1 3 0 7 : 1 5 0 7 : 3 5 0 7 : 2 4 0 7 : 2 6 0 7 : 4 6 0
6/7=8/9
5/7=7/9
7/8 = 8/9
6/8 = 7/9
abstract representationguarantees 7/9
abstract representationguarantees 7/9 abstract representation
guarantees 8/9
H
H0H1NEVER GOES DOWN: DERANDOMIZATION
UBCSAT
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10 : 1 010 : 2 010 : 3 0 7 : 1 2 0 7 : 1 3 0 7 : 2 3 0
rank 210: 1 = or(1) 7: 1 2 = or(!1 !2)
5 : 1 010 : 2 010 : 3 013 : 1 2 013 : 1 3 0 7 : 2 3 0
rank 25: 1 = or(!1)13: 1 2 = or(1 !2)
4/6 4/6
3/6 3/6
abstract representation guarantees0.625 * 6 = 3.75: 4 satisfied.
4/6 4/6
3/6 3/6
4/6 4/6
0 1 2 3
The effect of n-map
Evergreen game: G = {10,7}How do you choose aCSP(G)-instance to minimizepayoff? 0.618 …
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First Impression
• The abstract representation = look-ahead polynomials seems useful for guiding the search.
• The look-ahead polynomials give us averages: the guidance can be misleading because of outliers.
• But how can we compute the look-ahead polynomials? How do the polynomials help play the Evergreen(3,2) game?
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Where we are
• Introduction
• Look-forward
• Look-backward
• SPOT: how to use the look-ahead polynomials together with superresolution.
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Look Forward
• Why?– To make informed decisions– To play the Evergreen game
• How?– Abstract representation based on look-ahead
polynomials
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Look-ahead Polynomial(Intuition)
• The look-ahead polynomial computes the expected fraction of satisfied constraints among all random assignments that are produced with bias p.
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Consider an instance: 40 variables,1000 constraints (1in3)
1, … ,40
22: 6 7 9 0
22: 12 27 38 0
Abstract representation:reduce the instance tolook-ahead polynomial 3p(1-p)2 = B1,3(p) (Bernstein)
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1in3
0
0.1
0.2
0.3
0.4
0.5
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Coin bias (Probability of setting a variable to true)
Fra
ctio
n o
f co
nst
rain
ts t
hat
ar
e g
uar
ante
ed t
o b
e sa
tisf
ied
3p(1-p)2 for MAX-CSP({22})
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Look-ahead Polynomial(Definition)
• H is a CSP(G) instance.
• N is an arbitrary assignment.
• The look-ahead polynomial laH,N(p) computes the expected fraction of satisfied constraints of H when each variable in N is flipped with probability p.
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The general case MAX-CSP(G)
G = {R1, … }, tR(F) = fraction of constraints in F that use R.
x = p appSATR(x) over all R is a super set of the Bernstein polynomials (computer graphics, weighted sum of Bernstein polynomials)
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Rational Bezier Curves
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Bernstein Polynomials
http://graphics.idav.ucdavis.edu/education/CAGDNotes/Bernstein-Polynomials.pdf
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all the appSATR(x) polynomials
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Look-ahead Polynomial in Action
• Focus on purely mathematical question first
• Algorithmic solution will follow
• Mathematical question: Given a CSP(G) instance. For which fractions f is there always an assignment satisfying fraction f of the constraints? In which constraint systems is it impossible to satisfy many constraints?
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Remember?
MAX-CSP(G,f): Given a CSP(G) instance H expressed in n variables which
may assume only the values 0 or 1, find an assignment to the n variables which satisfies at least the fraction f of the constraints in H.
Example: G = {22} of rank 3MAX-CSP({22},f):
22:1 2 3 0 22:1 2 4 0 22:1 3 4 022: 2 3 4 0
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Mathematical Critical Transition Point
MAX-CSP({22},f):
For f ≤ u: problem has always a solutionFor f ≥ u + : problem has not always a solution,
u critical transition point
always (fluid)
not always (solid)
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The Magic Number
• u = 4/9
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1in3
0
0.1
0.2
0.3
0.4
0.5
0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0
Coin bias (Probability of setting a variable to true)
Fra
ctio
n o
f co
nst
rain
ts t
hat
ar
e g
uar
ante
ed t
o b
e sa
tisf
ied
3p(1-p)2 for MAX-CSP({22})
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Produce the Magic Number
• Use an optimally biased coin– 1/3 in this case
• In general: min max problem
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The 22 reductions:Needed for implementation
22 60
3
240
15 255
0
1,0
1,1
2,1
2,0
3,03,1
3,0
3,1
2,0
2,1
22 is expanded into 6 additionalrelations.
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The 22 N-Mappings:Needed for implementation
22
41
73
134
97
146
148
0
2
1
22 is expanded into 7 additionalrelations.
1040
21
0
2
1
0
2
1
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General Dichotomy Theorem
MAX-CSP(G,f): For each finite set G of relations closedunder renaming there exists an algebraic number tG :
For f ≤ tG: MAX-CSP(G,f) has polynomial solutionFor f ≥ tG+ : MAX-CSP(G,f) is NP-complete,
tG critical transition pointeasy (fluid)Polynomial
hard (solid)NP-complete
due to Lieberherr/Specker (1979, 1982)
polynomial solution:Use optimally biased coin.Derandomize.P-Optimal.
implications for the Evergreen game? Are you a better player?
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Context
• Ladner [Lad 75]: if P !=NP, then there are decision problems in NP that are neither NP-complete, nor they belong to P.
• Conceivable that MAX-CSP(G,f) contains problems of intermediate complexity.
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General Dichotomy Theorem(Discussion)
MAX-CSP(G,f): For each finite set G of relations closedunder renaming there exists an algebraic number tG
For f ≤ tG: MAX-CSP(G,f) has polynomial solutionFor f ≥ tG+ : MAX-CSP(G,f) is NP-complete,
tG critical transition pointeasy (fluid), Polynomial (finding an assignment)constant proofs (done statically using look-ahead polynomials)no clause learning
hard (solid), NP-completeexponential, super-polynomial proofs ???relies on clause learning
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min max problem
tG = min max sat(H,M)
all (0,1) assignments M
all CSP(G)instances H
sat(H,M) = fraction of satisfied constraints in CSP(G)-instance H by assignment M
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Problem reductions are the key
• Solution to simpler problem implies solution to original problem.
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min max problem
tG = lim min max sat(H,M,n)
all (0,1) assignments Mto n variables
n toinfinity
sat(H,M,n) = fraction of satisfied constraints inCSP(G)-instance H by assignment M with n variables.
all SYMMETRIC CSP(G) -instances H with n variables
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Reduction achieved
• Instead of minimizing over all constraint systems it is sufficient to minimize over the symmetric constraint systems.
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Reduction
• Symmetric case is the worst-case: If in a symmetric constraint system the fraction f of constraints can be satisfied, then in any constraint system the fraction f can be satisfied.
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Symmetric the worst-case
.
.
n variablesn! permutations
If in the big system thefraction f is satisfied, then there mustbe a least one small systemwhere the fraction f is satisfied
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min max problem
tG = lim min max sat(H,M,n)
all (0,1) assignments Mto n variables where thefirst k variables are setto 1
n toinfinity
sat(H,M,n) = fraction of satisfied constraints in system S by assignment I
all SYMMETRIC CSP(G) -instances H with n variables
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Observations
• The look-ahead polynomial look-forward approach has not been used in state-of-the-art MAX-SAT and Boolean MAX-CSP solvers.
• Often a fair coin is used. The optimally biased coin is often significantly better.
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The Game Evergreen(r,m) for Boolean MAX-CSP(G), r>1,m>0
Two players: They agree on a protocol P1 to choose a set of m relations of rank r.
1. The players use P1 to choose a set G of m relations of rank r.
2. Anna constructs a CSP(G) instance H with 1000 variables and at most 2*m*(1000 choose r) constraints and gives it to player 2 (1 second limit).
3. Bob gets paid the fraction of constraints she can satisfy in H (100 seconds limit).
4. Take turns (go to 1).
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ForEvergreen(3,2)
100% R1, 0% R2100% R2, 0% R1
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Evergreen(3,2) protocol possibilities
• Variant 1– Player Bob chooses both relations G – Player Anna chooses CSP(G) instance H.– Player Bob solves H and gets paid by Anna.
• This gives too much control to Bob. Bob can choose two odd relations which guarantees him a pay of 1 independent of how Anna chooses the instance H.
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Evergreen(3,2) protocol possibilities
• Variant 2:– Anna chooses a relation R1 (e.g. 22).– Bob chooses a relation R2.– Anna chooses CSP(G) instance H.– Bob solves H and gets paid by Anna.
R1, Anna R2, Bob
100% R1, 0% R2100% R2, 0% R1
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Problem with variant 2
• Anna can just ignore relation R2.
• Gives Anna too much control because the payoff for Bob depends only on R1 chosen by Anna (and the quality of the solver that Bob uses).
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Protocol choice: variant 3
• Randomly choose R1 and R2 (independently) between 1 and 255 (Throw two dice).
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Tell me
• How would you react as Anna?– The relations 22 and 22 have been chosen.– You must create a CSP({22}) instance with
1000 variables in which only the smallest possible fraction can be satisfied.
– What kind of instance will this be? 4/9
• What kind of algorithm should P1 use to maximize its payoff? compute optimal k + best MAX-CSP solver.
symmetric instance with (1000 choose 3) constraints
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ForEvergreen(3,2)
100% R1, 0% R2100% R2, 0% R1
Tells us how tomix the tworelations
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Role of tG in the Evergreen(3,2) game
• 1: Instance construction: Choose a CSP(G) instance so that only the fraction tG can be satisfied: symmetric formula.
• 2: Choose an algorithm so that at least the fraction tG is satisfied. (2 gets paid tG from 1).
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Game strategy in a nutshell
• Anna: Best: Choose tG instance
• Bob: Get’s paid tG
• etc.
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Additional Information
• Rich literature on clause learning in SAT and CSP solver domain. Superresolution is the most general form of clause learning with restarts.
• Papers on look-ahead polynomials and superresolution: http://www.ccs.neu.edu/research/demeter/papers/publications.html
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Additional Information
• Useful unpublished paper on look-ahead polynomials: http://www.ccs.neu.edu/research/demeter/biblio/partial-sat-II.html
• Technical report on the topic of this talk: http://www.ccs.neu.edu/research/demeter/biblio/POptMAXCSP.html
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Future work
• Exploring best combination of look-forward and look-back techniques.
• Find all maximum-assignments or estimate their number.
• Robustness of maximum assignments.
• Are our MAX-CSP solvers useful for reasoning about biological pathways?
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Conclusions
• Presented SPOT, a family of MAX-CSP solvers based on look-ahead polynomials and non-chronological backtracking.
• SPOT has a desirable property: P-optimal.• SPOT can be implemented very efficiently.• Preliminary experimental results are
encouraging. A lot more work is needed to assess the practical value of the look-ahead polynomials.
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Polynomials for rank 3
• x^3 x^2 x^1 x^0 relation• -1 3 -3 1 1• 1 -2 1 0 2 • 0 1 -2 1 3• 1 -2 1 0 4• 0 1 -2 1 5For 2: x*(1-x)2=x3-2x2+xmaximum at x=1/3; 1/3*(2/3)2=4/27
Check: 2 and 4 are the same
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Polynomials for rank 3
• x^3 x^2 x^1 x^0 relation• -1 3 -3 1 1• 1 -2 1 0 2 • 0 1 -2 1 3• 1 -2 1 0 4 (same as 2)• 0 1 -2 1 5For 4: x*(1-x)2=x3-2x2+xmaximum at x=1/3; 1/3*(2/3)2=4/27
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Recall
• (f*g)’ = f’*g + f*g’
• (f2)’ = 2*f * f’
• For relation 2:– x*(1-x)2 = (1-x)2 + x*2(1-x)*(-1)= (1-x)(1-3x)– x=1 is minimum– x=1/3 is maximum– value at maximum: 4/27
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Harold• concern: intension, extension: query, predicate
• extension = intension(software)
• Harold Ossher: confirmed pointcuts
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The Game Evergreen(r,m) for Boolean MAX-CSP(G), r>1,m>0
Two players: They agree on a protocol P1 to choose a set of m relations of rank r.
1. The players use P1 to choose a set G of m relations of rank r.
2. Anna constructs a CSP(G) instance H with 1000 variables and at most 2*m*(1000 choose r) constraints and gives it to Bob (1 second limit).
3. Bob gets paid by Anna the fraction of constraints he can satisfy in H (100 seconds limit).
4. Take turns (go to 1).
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Evergreen(3,2)
• Rank 3: Represent relations by the integer corresponding to the truth table in standard sorted order 000 – 111.
• choose relations between 1 and 254 (exclude 0 and 255).
• Don’t choose two odd numbers: All false would satisfy all constraints.
• Don’t choose both numbers above 128: All true would satisfy all constraints.
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How to play Evergreen(3,2)
• G = {R1, R2} is given (by some protocol)
• Anna: compute t=(t1, t2) so that
max appmeant(x) for x in [0,1] is minimum.
Construct a symmetric instance SYMG(t) H.
• Bob: Solves H.
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Question
• For any G and any CSP(G)-instance H, is there a weight assignment to the constraints of H so that the look-ahead polynomial absH has its maximum not at 0 or 1 and guarantees a maximum assignment for H without weights? – the polynomial might guarantee maximum-
1+e which is enough to guarantee a maximum assignment.
– what if we also allow n-maps?
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Absolute P-optimality
• Bringing max to boundary is polynomial.• Bringing max away from boundary using weights? What
is the complexity.• Definition: ImproveLookAhead(G,H,N): Given G, a
CSP(instance) H and an assignment N for H. Is there an assignment that satisfies at least laH,N(mb) + 1. mb = maximum bias.
• Assume G sufficiently closed. • Theorem: [Absolute P-optimality]
ImproveLookAhead(G,H,N) is NP-hard iff MAX-CSP(G) is NP-hard.
• Warning: ImproveAllZero(G,H) is NP-hard iff MAX-CSP(G) is NP-hard.
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Exploring the search space
• Look-ahead polynomials don’t eliminate parts of the search space.
• They crosscut the search space early in the search process. Whenever the look-ahead polynomial guarantees more than the currently best assignment, we can cut across the search space but might have to get back to the part we jumped over.
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Crosscutting the search spacecurrent
better
best
by look-ahead
evenbetter
by search
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Early better than later
• Look-ahead polynomials are more useful early in the search.
• Later in the search the maximum will be at 0 or 1.
• Look-ahead polynomials will make mistakes which are compensated by superresolvents.
• Superresolvents cut part of the search space and they help the look-ahead polynomials to eliminate the mistakes.
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Requirements for algorithms and properties to work
• Relative P-optimality
• Absolute P-optimality– G needs to be closed under renaming and
reductions and n-maps
• Look-ahead polynomials– improve assignments: closed under n-maps
and reductions
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Never require closed under renaming?
• symmetric formulas don’t require it? They do? Consider
2: 1 2 3 0
2: 1 2 4 0
2: 1 3 4 0
2: 2 3 4 0
is not symmetric. {1 !2 !3 4} does not satisfy all, only ¾. {!1 2 3 !4} only satisfies ¼.
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What happens during the solution process
• Maximum of polynomial will be at the boundary, say 0. Can be achieved in P. Notice folding effect.
• Many superresolvents will be learned until better assignment is found.
• Most constraints use an odd relation, a few an even relation (if many constraints can be satisfied).
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What happens …
• Because the polynomial only depends on a few numbers, it is not sensitive to the detailed properties of the instance.
• But if one variable has a visible bias towards either 1 or 0, polynomials might detect it.
• Adjust the weight of the constraints to bring the maximum of the polynomial into the middle so that abs(mb) increases.
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Question for Daniel
• p(x) = t1*p1(x)+t2*p2(x)
• mb at 0
• p(mb)
• perturb t1,t2 so that p(x) gets a higher maximum. The fraction of t1 should go up if R1 is an unsatisfied relation.
• How high can we bring the fraction of satisfied constraints this way?
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Question
• Does this solve the original problem?
• If we get all satisfied, yes.
• Can force that, by deleting all but one unsatisfied and adding them later on???
• Are forced to work with many relations.
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SAT Rank 2 instance
14 : 1 2 014 : 3 4 014 : 5 6 0 7 : 1 3 0 7 : 1 5 0 7 : 3 5 0 7 : 2 4 0 7 : 2 6 0 7 : 4 6 0
14: 1 2 = or(1 2) 7: 1 3 = or(!1 !3)
= F
find maximum assignmentand proof that it is maximum
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Solution Strategy
• The MAX-CSP transition system gives many options:– Choose initial assignment. Has significant
impact on length of proof. Best to start with a maximum assignment.
– variable ordering. Irrelevant because start with maximum assignment.
– value ordering: Also irrelevant.
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SAT Rank 2 instance
14 : 1 2 014 : 3 4 014 : 5 6 0 7 : 1 3 0 7 : 1 5 0 7 : 3 5 0 7 : 2 4 0 7 : 2 6 0 7 : 4 6 0
14: 1 2 = or(1 2) 7: 1 3 = or(!1 !3)
N={1 !2 !3 4 5 !6}unsat=1/9{}|F|{}|N -> D UP*{1* !3 !5 4 6}|F|{}|N -> SSR Restart{}|F|5(1)|N -> UP*{!1 2 !4 !6 5 3}F|5(1)|N -> SSR{!1 2 !4 !6 5 3}F|5(1),0()|N -> Finaleend
rank 210: 1 = or(1) 5: 1 = or(!1)
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Rank 2 relations
ba1 00 0 02 01 1 04 10 0 18 11 1 1 10 1210(1) = or(1) = or(*,1), don’t mention second
argument12(1) = or(1) = or(1,*), 10(2,1) = 12(1,2)0() = empty clause
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UP / D
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Variable ordering
• maximizes likelihood that look-ahead polynomials make correct decisions
• Finds variable where look-ahead polynomials give the strongest indication– even if the look-ahead polynomial chooses
the wrong mb, the decision might still be right– what is better:
• laH1(mb1) is max• | laH1(mb1) - laH0(mb0) | is max (is more instance
specific. Will adapt to superresolvents.)
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mean versus appmean
• mean does less averaging, so it is preferred?
• appmean looks at the neighborhood of x*n
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Derandomization
• In a perfectly symmetric CSP(G) instance, it is sufficient to try any assignment with k ones for k from 0 to n to achieve the maximum (tG).
• But in a non-symmetric instance, we need derandomization to achieve tG and superresolution to achieve the maximum.
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The Game EvergreenTM(r,m) for Boolean MAX-CSP(G), r>1,m>0
Two players: They agree on a protocol P1 to choose a set of m relations of rank r.
1. The players use P1 to choose a set G of m relations of rank r.2. Anna constructs a CSP(G) instance H with 1000 variables and
at most 2*m*(1000 choose r) constraints and gives it to Bob (1 second limit). Anna knows the maximum assignment and has a proof for maximality but keeps it secret until Bob gives response.
3. Bob gets paid the fraction of constraints he can satisfy in H relative to the maximum number that can be satisfied (100 seconds limit).
4. Take turns (go to 1).
TM = true maximum
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EvergreenTM versus Evergreen
• Anna can try to create instances that are hard to solve by Bob’s solver.
• If Bob has a perfect solver, he will be paid 1.0.
• The game depends a lot on the solver quality.
• Incomplete information (maximum assignment is kept secret).
• Challenge for Anna to find instance where maximum is known with short proof.
• Anna can control the maximum Bob is paid assuming a perfect solver.
• Bob may be paid little even with a perfect solver.
• The game depends less on solver quality.
• Complete information.
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Using Mathematica
• Combine2[15, 238]– t1(1-x)-t2(-2+x)*x
• D[D[Combine2[15, 238],x],x]– -2*t2 is negative: must be a maximum
• Solve[D[Combine2[15, 238],x]=0,x]– x = -t1+2*t2/2*t2
• RootsOf2[15,238]/.t2->(1-t1)/.t1->1/5(5-sqrt(5)
15: 1 0 = !1238: 1 2 0 = 1 or 2
UBC March 2007 91
Mathematica
• Solve2(15, 238)– ½ (sqrt(5)-1)– t1 = 1-1/sqrt(5)– t2 = 1/sqrt(5)
• Solve2(22, 22)– 4/9– t1 = ½– t2 = ½
UBC March 2007 92
MathematicaIncludeIt[r_, n_] := Mod[Floor[r/n], 2];
AppSAT[r_] := Simplify[IncludeIt[r, 1]*x^0*((1 - x))^((3 - 0)) + (( IncludeIt[r, 2] + IncludeIt[r, 4] + IncludeIt[r, 16]))* x^1*((1 - x))^((3 - 1)) + ((IncludeIt[r, 8] + IncludeIt[r, 32] + IncludeIt[r, 64]))* x^2*((1 - x))^((3 - 2)) + IncludeIt[r, 128]*x^3*((1 - x))^((3 - 3))];Combine2[r1_, r2_] := t1*AppSAT[r1] + t2*AppSAT[r2];RootsOf2[r1_, r2_] := ReplaceAll[Combine2[r1, r2], Solve[D[Combine2[r1, r2], x] == 0, x]];Solve2[r1_, r2_] := For [i = 1, i <= Length[ RootsOf2[r1, r2]], i++, Print[Minimize[{ RootsOf2[r1, r2][[ i]], 0 < t1 < 1, 0 < t2 < 1, t1 + t2 == 1}, {t1, t2}]]];